# Volume of Gabriel's horn using double integral

I am trying to use a double integral to calculate the volume of Gabriel's horn:

https://en.wikipedia.org/wiki/Gabriel%27s_Horn

$V = \int\int_R (x^2+y^2)^{-1/2} dx dy$

Converting to polar coordinates:

$V = \int_{0}^{2\pi} \int_{0}^{1} \frac{1}{r} r\:dr\:d\theta$

$=2\pi$

But the answer should be $V = \pi$.

Should the limits for $\theta$ be $0 \rightarrow \pi$ or have I made a mistake somewhere else?

I think the integral you want is $$\int\int_R\left(\frac1{\sqrt{x^2+y^2}}-1\right)\,dx\,dy$$ where $R$ is the unit disc. This gives the volume of the set $$\{(x,y,z):z\sqrt{x^2+y^2}\le1,z\ge1\}.$$

• But the equation of the surface is $f(x,y) = \frac{1}{\sqrt{x^2+y^2}}$. Where does the $-1$ come from? – 1123581321 Jun 18 '17 at 6:00

Gabriel's horn is obtained by rotating the graph of $f(x)=\frac{1}{x}$ over $(1,+\infty)$ around the $x$-axis.
For any $t\in(1,+\infty)$ the section $x=t$ is given by a circle with radius $\frac{1}{t}$ and area $\frac{\pi}{t^2}$, hence by Cavalieri's principle $$V = \int_{1}^{+\infty}\frac{\pi}{t^2}\,dt = \color{red}{\large\pi}.$$

The problem with your analysis is that the volume should be expressed formally as

$$V=2\pi\int_1^{\infty}\int_0^{1/x}y\,dy~dx=2\pi\int_1^{\infty}\frac{y^2}{2}\biggr|_0^{1/x}~dx=\pi\int_1^{\infty}\frac{1}{x^2}~dx=\pi$$

• Why is the integrand $y$ in the first integral? – 1123581321 Jun 19 '17 at 8:26
• $2\pi y dy dx$ is the elemental volume being integrated; first along $y$ to the the area of a disc normal to $x$-axis then along $x$ to get the volume of the horn. – Cye Waldman Jun 19 '17 at 13:24